What is the solution for (y-2x-1)dx+(x+y-4)dy=0 ?

The solution for (y-2x-1)dx+(x+y-4)dy=0 is y=4-x+sqrt(3x^2-6x+c_{1)+16},y=4-x-sqrt(3x^2-6x+c_{1)+16}

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(y-2x-1)dx+(x+y-4)dy=0

(y - 2 x - 1) d x + (x + y - 4) d y = 0

y = 4 - x + 3 x^{2} - 6 x + c_{1} + 16, y = 4 - x - 3 x^{2} - 6 x + c_{1} + 16

Solution steps

(y - 2 x - 1) d x + (x + y - 4) d y = 0

Solve exact ODE:

y = 4 - x + 3 x^{2} - 6 x + c_{1} + 16, y = 4 - x - 3 x^{2} - 6 x + c_{1} + 16

y = 4 - x + 3 x^{2} - 6 x + c_{1} + 16, y = 4 - x - 3 x^{2} - 6 x + c_{1} + 16

n \to \infty lim (4^{n})

x = - 2, x = 3, y = 2 x^{2} + 10, y = 0

f (x) = 9 e^{x} cos (x), at x = 0

\frac{\partial}{\partial v} (4 u cos (v))

\frac{\partial}{\partial x} (\frac{1}{1 + e ^{- x}})

What is the solution for (y-2x-1)dx+(x+y-4)dy=0 ?
The solution for (y-2x-1)dx+(x+y-4)dy=0 is y=4-x+sqrt(3x^2-6x+c_{1)+16},y=4-x-sqrt(3x^2-6x+c_{1)+16}